Evaluate the definite integrals by expanding the integrand in a Maclaurin series. \(\int_{0}^{0.1} e^{-x^{2}} d x\)

The Maclaurin series is a special case of the Taylor series, centered at zero. It is an infinite series used to represent functions. The general form of the Maclaurin series for a function, \( f(x) \), is given by:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \]
The exponential function, \( e^x \), is particularly nice because it has an easily recognizable Maclaurin series:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
In the exercise, we use the Maclaurin series to approximate the function \( e^{-x^2} \), by substituting \( -x^2 \) for \( x \):
\[ e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots \]
This substitution helps convert the complex function into an infinite polynomial series, making it easier to integrate term-by-term.

A definite integral calculates the area under a curve within specified limits. For a function \( f(x) \) integrated from \( a \) to \( b \), it is expressed as:
\[ \int_a^b f(x) \, dx \]
In our exercise, we are tasked with evaluating the definite integral:
\[ \int_{0}^{0.1} e^{-x^2} \, dx \]
This integral calculates the area under the curve \(e^{-x^2}\) from \( x = 0 \) to \( x = 0.1 \). Due to the complexity of the function, we expand \( e^{-x^2} \) using its Maclaurin series before integrating.

The term-by-term integration involves integrating each term of a series individually. This method simplifies integration of complex or non-standard functions.
In our exercise, we substitute the Maclaurin series for \( e^{-x^2} \) into the integral:
\[ \int_{0}^{0.1} e^{-x^2} \, dx = \int_{0}^{0.1} \left(1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots\right) \, dx \]
We integrate each term separately:

• \( \int_{0}^{0.1} 1 \, dx = 0.1 \)
• \( \int_{0}^{0.1} (-x^2) \, dx = -\frac{(0.1)^3}{3} = -0.000333 \)
• \( \int_{0}^{0.1} \frac{x^4}{2!} \, dx = \frac{(0.1)^5}{10} = 0.000001 \)

Higher-order terms become extremely small and are often neglected, simplifying our calculations further.

Exponential functions involve the constant \( e \) (approximately 2.718). They are critical in various fields like mathematics, science, and engineering.
The general form of an exponential function is \( e^x \). When combined with powers or exponentials of variables, these functions become more complex to integrate.
In the context of our integral:
\[ e^{-x^2} \]
We encounter an exponential function with a negative quadratic exponent. This makes direct integration challenging, but approximations like the Maclaurin series help simplify the process. By expanding \( e^{-x^2} \) into a series, we can perform term-by-term integration to find:
\[ \int_{0}^{0.1} e^{-x^2} \, dx \approx 0.099668 \]
This approach is a powerful tool for working with exponential functions where exact integrals are not easily attainable.